
Here, B=0.75T
The magnitude of magnetic force on the 5cm side is {F}_{AB}=i\times ({l}_{AB})B\mathrm{sin}\theta (\text{angle between}AC&AB)
=2×(1005)×0.75×1312
=1309N
Hence, the value of x=9.
JEE Main 2023 — Physics Electromagnetism
A single turn current loop in the shape of a right angle triangle with sides 5cm,12cm,13cm is carrying a current of 2A. The loop is in a uniform magnetic field of magnitude 0.75T whose direction is parallel to the current in the 13cm side of the loop. The magnitude of the magnetic force on the 5cm side will be 130xN. The value of x is _______.
Held on 24 Jan 2023 · Verified 6 Jul 2026.
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